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Question Number 34219 by abdo imad last updated on 03/May/18

$$calculate{\int }_0^{\frac{\pi }{4}}\frac{dx}{{cos}^{3}x+{sin}^{3}x}$$

Commented by abdo imad last updated on 31/May/18

$$I={\int }_0^{\frac{\pi }{4}}\frac{dx}{(cosx+sinx)({cos}^{2}x-cosxsinx+{sin}^{2}x)}$$$$={\int }_0^{\frac{\pi }{4}}\frac{dx}{(cosx+sinx)(1-cosxsinx)}$$$$={\int }_0^{\frac{\pi }{2}}\frac{dx}{\sqrt{2}cos(\frac{\pi }{4}-x)(1-\frac{1}{2}sin\left(2x\right))}$$$${=}_{\frac{\pi }{4}-x=t}{\int }_{\frac{\pi }{4}}^{-\frac{\pi }{4}}\frac{-dt}{\sqrt{2}cost(1-\frac{1}{2}sin\left(2(\frac{\pi }{4}-t)\right)}$$$$=\sqrt{2}{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{dt}{cos\left(t\right)(2-sin(\frac{\pi }{2}-2t))}$$$$=\sqrt{2}{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{dt}{cos\left(t\right)(2-cos\left(2t\right))}$$$$=2\sqrt{2}{\int }_0^{\frac{\pi }{4}}\frac{dt}{cost(2-2{cos}^{2}t+1)}$$$$=2\sqrt{2}{\int }_0^{\frac{\pi }{4}}\frac{dt}{cost(3-2{cos}^{2}t)}changementtan\left(\frac{t}{2}\right)=x$$$$give$$$$I=2\sqrt{2}{\int }_0^{\sqrt{2}-1}\frac{1}{\frac{1-x^2}{1+x^2}\{3-2{\left(\frac{1-x^2}{1+x^2}\right)}^2\}}\frac{2dx}{1+x^2}$$$$=4\sqrt{2}{\int }_0^{\sqrt{2}-1}\frac{dx}{(1-x^2)\left\{\frac{3{(1+x^2)}^2-2{(1-x^2)}^2}{{(1+x^2)}^2}\right\}}$$$$=4\sqrt{2}{\int }_0^{\sqrt{2}-1}\frac{{(1+x^2)}^2}{(1-x^2)\{3{(1+x^2)}^2-2{(1-x^2)}^2\}}dx$$$$=4\sqrt{2}{\int }_0^{\sqrt{2}-1}\frac{x^4+2x^2+1}{(1-x^2)(3x^4+6x^2+3-2x^4+4x^2-2\}}dx$$$$=4\sqrt{2}{\int }_0^{\sqrt{2}-1}\frac{x^4+2x^2+1}{(1-x^2)(x^4+10x^2+1)}dx....becontinued...$$

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